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Answer 1

. it can be observed that vertical asymptotes exist at x = -3 8 2= 4 . This suggests that denominator consists of the factors ( 2 + 3 ) & ( 21 -4) . it can be observed that the graph cuts the 2 - axis at 2 = 2 . Thus , numerator has a factor of ( 2 - 2 ) . Let f ( x ) = 1 (2 - 2) (2+ 3) ( 2 - 4 ) Now horizontal asymptote exists at $= 0 or it is y anis . This suggests lim f ( 2 ) = 0 lim * ( 2 - 2 ) = 0 ( 2 + 3 ) ( 22 - 4 ) lim Rx ( 1 - 2/x) =0 2 / 1 + 3 lim k (1- 2/ 2) = 0 x (1+3 /26) (1 - 4/21) On substituting 2=0. we get that I can take any value\f\f\fStep 1 of 2:) Introduction: A matrix A is invertible if def ( A ) 7 0 Step 2 of 2:) Solution : Here A is a 3x 3 matrix such that ( A + I3 ) 2 = 0 -> ( A + 13 ) ( A + [3 ) = 0 => A . A + A . [3 + I; A + 13. 13 = 0 = > A 2 + A + A + 13 =0 -> A2 + 2 A = - I3 -> A ( A + 2 I3 ) = - Iz del SA ( A + 2 Is ) ] = det ( - 13 ) - det ( A ) det ( A+ 25 ; ) = ( 1) 3 ded Is -> ded ( D ) det ( A + 213 ) = - 1 ; [: del 13 = 1 ] -7 def ( A ) # 0 Hence A is invertible ( proved )13 . Let a random no. represents if that through was successful. We take 10 Shots ( 10 random no . ) as our paperiment and see if we with A win means we have more than 65 %% probability of experiment Experiment No of distinct shots Wlin ( y / N ) 52711.38889 ( 1 1 2 , 3 , 5 , 7 , 8 , 9 ) = 7 y ( 7 / 10 ) 2 . g 30 74 60 227 ( 2 , 3 , 4 1 6 , 7 1 9,0 ) = 7 y ( 7 / 10 ) 3 . 4001 185848 (1 4 , 5 , 8, 0) = 5 N ( 5 / 10 ) 4. 48 76 75 2573 ( 2 3 , 4 , 5 , 6 , 7, 8 ) = 7 y ( 7/ 10 ) Probability = 3/ 4 = 0.75 ( 3 success full experiments out of 4 )